Simplify the following expression and state the condition under which the simplification is valid. You can assume that $y \neq 0$. $r = \dfrac{-4}{35y - 21} \times \dfrac{3(5y - 3)}{4y} $
Answer: When multiplying fractions, we multiply the numerators and the denominators. $r = \dfrac{ -4 \times 3(5y - 3) } { (35y - 21) \times 4y } $ $ r = \dfrac {-4 \times 3(5y - 3)} {4y \times 7(5y - 3)} $ $ r = \dfrac{-12(5y - 3)}{28y(5y - 3)} $ We can cancel the $5y - 3$ so long as $5y - 3 \neq 0$ Therefore $y \neq \dfrac{3}{5}$ $r = \dfrac{-12 \cancel{(5y - 3})}{28y \cancel{(5y - 3)}} = -\dfrac{12}{28y} = -\dfrac{3}{7y} $